The Question is from neet-physics atoms-nuclei , and it was created by the author VENKATESWARARAO

Question

Given the value of Rydberg constants is ${\rm{1}}{{\rm{0}}^7}\,{{\rm{m}}^{ - 1}}$, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Rydberg Formula and Balmer Series
	**Related Topics**: Atomic Spectra, Quantum Numbers, Electronic Transitions
	**Prerequisites**: Understanding of Atomic Structure and Spectroscopy
## SOLUTION PATHWAY
**Given Information**:
	* Rydberg constant (R) = $10^7 m^{-1}$
	* Balmer series: transitions to n=2
	* Last line of Balmer series corresponds to transition from n = ∞ to n = 2
**Method & Steps**:
	1.  Identify the relevant formula: The Rydberg formula for wave number ($\overline{v}$) is given by:  $\overline{v} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$
		Reasoning: This formula calculates the wave number of emitted light during electron transitions in a hydrogen atom.
		Formula/Rule: $\overline{v} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$
	2. Apply the formula for the last line of Balmer series: For the last line of the Balmer series, $n_1 = 2$ and $n_2 = \infty$. Substituting these values, we get: $\overline{v} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight)$
		Working: $\overline{v} = 10^7 \left( \frac{1}{4} - 0 ight) = 10^7 	imes \frac{1}{4}$
		Key Point:  $\frac{1}{\infty^2}$ approaches 0.
	3. Calculate the wave number:  $\overline{v} = 0.25 	imes 10^7 m^{-1}$
		Result: The wave number is $0.25 	imes 10^7 m^{-1}$
		Verification:  The result is in the expected order of magnitude for the wave number in the Balmer series.
 ## OPTION ANALYSIS
[A] :x: This value is incorrect.  It likely arises from a misunderstanding of the formula or an arithmetic error.
[B] :x: This value is incorrect.  It might be the Rydberg constant itself or a miscalculation.
[C] ✓: This is the correct wave number calculated based on the Rydberg formula for the last line of the Balmer series.
[D] :x: This value is incorrect. It is a multiple of the correct answer.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. [Error 1]: Using the wrong values for $n_1$ and $n_2$ for the Balmer series. [How to avoid]: Remember that the Balmer series corresponds to transitions to $n=2$, and for the last line, the electron comes from $n=\infty$.
		2. [Error 2]:  Incorrectly calculating the value of the wave number. [How to avoid]: Double-check your arithmetic and ensure you are using the correct formula.
	**Quick Tips**:
		1. [Helpful shortcut]:  Remember that for the last line of any series, the second term in the Rydberg formula becomes zero.
		2. [Memory aid]: The Rydberg formula is key to understanding atomic spectra; memorize it.

VENKATESWARARAO · Answer

${m{0}}{m{.025}}\, 	imes {m{1}}{{m{0}}^4}\,{{m{m}}^{ - 1}}\,$

VENKATESWARARAO · Answer

${m{0}}{m{.5}}\, 	imes {m{1}}{{m{0}}^7}\,{{m{m}}^{ - 1}}$

VENKATESWARARAO · Answer

The correct Option is "${m{0}}{m{.25}}\, 	imes {m{1}}{{m{0}}^7}\,{{m{m}}^{ - 1}}$"

VENKATESWARARAO · Answer

${m{2}}{m{.5}}\, 	imes {m{1}}{{m{0}}^7}\,{{m{m}}^{ - 1}}$

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